OFFSET
1,2
COMMENTS
Multiplicative, because products and quotients of multiplicative functions are always multiplicative. a(p^k) for fixed k is trivially a rational function of p. The proofs below show that a(n) is always an integer, and hence a(p^k) is a polynomial in p. (Note that a(n^k) is not equal to this polynomial when n is composite.)
Proof from Robert Gerbicz that this is always an integer:
First consider r=sigma_2(p^k)/sigma(p^k). Let x=p^k, then
r=(p^(2*k+2)-1)/(p^2-1)*(p-1)/(p^(k+1)-1)=(p^2*x^2-1)/(p^2-1)*(p-1)/(p*x-1)=(p*x+1)/(p+1)
Case a: k is even, then x==1 mod (p+1), so (p*x+1)==-1+1==0 mod (p+1). So here even r is an integer.
Case b: k is odd, then
sigma_3(p^k)=(p^3*x^3-1)/(p^3-1), and we must multiple this by r, and here
p^3*x^3-1==0 mod (p+1)
p^3-1==-2 mod (p+1)
This means that if p=2 then sigma_3(p^k) is divisible by (p+1), so we have proved the theorem.
If p is odd, then (p+1)/2 divides sigma_3(p^k), but we have another factor of 2 in (p*x+1), because p and x is odd.
It appears that a stronger result is also true: sigma_3(p^k) is divisible by (p+1) if k is odd.
Proof from Giovanni Resta that this is always an integer:
We have sigma_2(p^k)*sigma_3(p^k)/sigma(p^k) =
((p^(2+2k)-1)/(p^2-1) * (p^(3+3k)-1)/(p^3-1)) / ((p^(1+k)-1)/(p-1)) =
((p^(k+1)+1) (p^(3 k+3)-1)) / ((p+1) (p^3-1)),
because p^(2+2k)-1 is the difference of two squares and we can use x^2-1 = (x+1)(x-1).
We observe that (p^(3 k+3)-1)) is always divisible by (p^3-1) (it is indeed sigma_3(p^k)).
Now, if k is even, k+1 is odd, so p^(k+1)+1 is divisible by p+1 giving 1-p+p^2-p^3... +p^k as result, and we are done.
If k is odd, k+1 is even and in general p^(k+1)+1 is not divisible by p+1 (just as p^2+1 is not divisible by p+1).
However, if k is odd, then 3k+3 is even, so p^(3k+3)-1 beside being divisible by p^3-1 is also divisible by p+1. Since p+1 and p^3-1 have no common factors, then the ratio (p^(3 k+3)-1)) / ((p+1) (p^3-1)) is an integer and we are done.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
FORMULA
a(n) = sigma_2(n)*sigma_3(n)/sigma(n).
Sum_{k=1..n} a(k) ~ c * n^5, where c = (Pi^6*zeta(5)/2700) * Product_{p prime} (1 - 2/p^2 + 2/p^3 - 2/p^4 + 1/p^6) = 0.1662831668... . - Amiram Eldar, Dec 01 2022
MATHEMATICA
a[n_] := DivisorSigma[2, n] * DivisorSigma[3, n] / DivisorSigma[1, n]; Array[a, 40] (* Amiram Eldar, Aug 01 2019 *)
(#[[2]]#[[3]])/#[[1]]&/@Table[DivisorSigma[k, n], {n, 40}, {k, 3}] (* Harvey P. Dale, Aug 14 2024 *)
PROG
(PARI) a(n) = sigma(n, 2)*sigma(n, 3)/sigma(n)
CROSSREFS
KEYWORD
nonn,mult
AUTHOR
Franklin T. Adams-Watters, Oct 24 2018
STATUS
approved